Whakaoti i te 1-t^2=1(t-t)

$$1 - t ^ { 2 } = 1 ( t - t )$$

Whakaoti mō t

$$t=1$$

$$t=-1$$

Ngā Upane mā te Kimi Pūtakerua

Pahekotia te t me -t, ka 0.

$$1-t^{2}=1\times 0$$

Whakareatia te 1 ki te 0, ka 0.

$$1-t^{2}=0$$

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-t^{2}=-1$$

Whakawehea ngā taha e rua ki te -1.

$$t^{2}=\frac{-1}{-1}$$

Whakawehea te -1 ki te -1, kia riro ko 1.

$$t^{2}=1$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$t=1 t=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Pahekotia te t me -t, ka 0.

$$1-t^{2}=1\times 0$$

Whakareatia te 1 ki te 0, ka 0.

$$1-t^{2}=0$$

Ko ngā tikanga tātai pūrua pēnei i tēnei nā, me te kīanga tau x^{2} engari kāore he kīanga tau x, ka taea tonu te whakaoti mā te whakamahi i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, ina tuhia ki te tānga ngahuru: ax^{2}+bx+c=0.

$$-t^{2}+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, 0 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{0±\sqrt{0^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Pūrua 0.

$$t=\frac{0±\sqrt{-4\left(-1\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$t=\frac{0±\sqrt{4}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 4.

$$t=\frac{0±2}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$t=\frac{0±2}{-2}$$

Nā, me whakaoti te whārite t=\frac{0±2}{-2} ina he tāpiri te ±. Whakawehe 2 ki te -2.

$$t=-1$$

Nā, me whakaoti te whārite t=\frac{0±2}{-2} ina he tango te ±. Whakawehe -2 ki te -2.

$$t=1$$

Kua oti te whārite te whakatau.

$$t=-1 t=1$$